Question

show that x=ct is an invariant under lorentz transformation​

Answer 1

Answer:

First, I have to assume that -y^2 is missing by accident. Second, you haven’t specified which Lorentz Transform. There are three perpendicular boosts in 4 dimensions. But the effects of a Lorentz Transform apply to a single space unit parallel to the path of the inertial frame. So, if we switch to Spherical coordinates, there is a single radius vector and two arbitrary angles. We are always free to rotate the coordinate system such that the radius vector is indeed parallel to the path. In this way, we don’t need to choose a Lorentz matrix, because Spherical coordinates eliminates two of the Cartesian coordinates (since they are invariant anyway), and there is only 1 2x2 Lorentz Transform for spacetime. Since we are only asked to show the invariance, not prove it, the simplest approach is to take an arbitrary event at (ct,r) and project it through an arbitrary Lorentz Transform, which is best represented parameterized in terms of arbitrary rapidity, w. Then, just do the arithmetic to see if (ct’,r’) has the same invariant.

The expression c²t²-r² = c²t²-x²-y²-z², so if c²t²-r² = c²t’²-r’², then the original question is answered. In terms of w,

│cosh(w) -sinh(w)││ct│ │ct’│

│-sinh(w) cosh(w)││ r│=│ r’│

Thus, ct’ = ct cosh(w) - r sinh(w), and r’ = r cosh(w) - ct sinh(w).

And ct’² = c²t² cosh²(w) - 2 r ct sinh(w)cosh(w) + r² sinh²(w),

while r’² = r² cosh²(w) - 2 r ct sinh(w)cosh(w) + c²t² sinh²(w)

ct’² - r’² = (c²t²-r²)cosh²(w) - (c²t²-r²)sinh²(w)

= (c²t²-r²)(cosh²(w)-sinh²(w)) = (c²t²-r²)

For an arbitrary event and an arbitrary Lorentz boost, ct’² - r’² =

ct² - r², the definition of invariance. QED